Find the \( z \)-score corresponding to the given area. Round to two decimals, if necessary, The area to the left of \( z \) is \( 60 \% \). \( z=\square \) The area to the right of \( z \) is \( 10 \% \). The area between \( -z \) and \( z \) is \( 97 \% \). \( z=\square \)

To find the \( z \)-score corresponding to the area to the left of \( z \) being \( 60\% \), you can use standard normal distribution tables or a calculator. The area of \( 60\% \) implies that \( z \) is the value where \( 60\% \) of the data lies to the left. Looking it up, the \( z \) corresponding to \( 0.60 \) is approximately \( 0.25 \). Next, for the area to the right of \( z \) being \( 10\% \), that leaves \( 90\% \) to the left. Therefore, you are looking for where \( 90\% \) of the distribution falls to the left. From the standard normal distribution table, the \( z \) corresponding to \( 0.90 \) is approximately \( 1.28 \). Lastly, for the area between \( -z \) and \( z \) being \( 97\% \), the area in the tails combined is \( 3\% \) (or \( 1.5\% \) in each tail). Therefore, you want to find the \( z \) value where \( 1.5\% \) is to the left. Look it up, and you find the corresponding \( z \) around \( 2.17 \). So, the answers are: 1. \( z \approx 0.25 \) 2. \( z \approx 1.28 \) 3. \( z \approx 2.17 \)